| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on smooth peg or cylinder |
| Difficulty | Challenging +1.2 This is a two-part moments problem requiring identification of the center of mass of a cone (given by formula at 3h/4 from vertex), setting up moment equations about pivot points, and resolving forces. Part (i) is straightforward equilibrium about V; part (ii) requires comparing limiting friction with toppling conditions. The geometry with the 60° angle adds some trigonometric work, but the problem follows standard M2 patterns for rigid body equilibrium with no particularly novel insights required. |
| Spec | 3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| i | \(F \times 0.8 = 0.6\cos 60 \times 550\) | M1 |
| \(F = 206.25\) | A1 | Accept 206, cao |
| A1 | ||
| A1 | ||
| [4] | ||
| ii | \(T \times 2 \times 0.8/\tan 30\) | A1 |
| \(= 550 \times (0.8/\sin 30 - 0.6\cos 60)\) | M1* | Moment of weight about P |
| \(T = 258\) | A1 | \(550 \times (1.6 - 0.3)\) Accept to 2sf |
| M1* | Resolving vertically, 3 terms needed | |
| \(R = 550 - T\cos 30\) | A1 | Value for \(T\) not required |
| \(Fr = T\sin 30\) | B1* | Value for \(T\) not required; accept \(<\) or \(\leq\) |
| \(\mu = 129/326.6\) | M1dep* | For correct use of \(F = \mu R, R \neq 550\) |
| \(\mu = 0.395\) | A1 | |
| [10] | ||
| OR | ||
| \(T \times 0.8/\tan 30 + 550 \times 0.6\cos 60 = R \times 0.8/\cos 60\) | M1* | Moments about V, 3 terms needed |
| \(R = 550 - T\cos 30\) | A2 | A1 for two terms correct |
| Solve for \(T\) or \(R\): \(T = 258\) or \(R = 326.5625\) | M1* | Resolving vertically, 3 terms needed |
| \(Fr = T\sin 30\) | B1* | accept \(<\) or \(\leq\) |
| \(\mu = 129/326.6\) | M1 | For correct use of \(F = \mu R, R \neq 550\) |
| \(\mu = 0.395\) | A1 | Only one needed. Accept to 2sf. For correct use of \(F = \mu R, R \neq 550\) |
| [10] | ||
| OR | ||
| \(Fr \times 1.6\cos 30 + 550 \times (1.6\sin 30 + 0.6\sin 30) = R \times (1.6 + 1.6\sin 30)\) | M1* | Moments about Q, 3 terms needed |
| \(R = 550 - T\cos 30\) | A2 | A1 for two terms correct |
| \(Fr = T\sin 30\) | B1* | accept \(<\) or \(\leq\) |
| Solving for at least one of \(R\), \(Fr\), or \(T\): Either \(R = 326.5625\), or \(Fr = 129(.0017008)\), or \(T = 258\) | M1 | |
| \(\mu = 129/326.6\) | M1dep* | |
| \(\mu = 0.395\) | A1 | |
| [10] |
| **Part** | **Answer/Working** | **Marks** | **Guidance** |
|----------|-------------------|----------|-------------|
| i | $F \times 0.8 = 0.6\cos 60 \times 550$ | M1 | Attempt at moments |
| | $F = 206.25$ | A1 | Accept 206, cao |
| | | A1 | |
| | | A1 | |
| | | [4] | |
| ii | $T \times 2 \times 0.8/\tan 30$ | A1 | Moment of $T$ about P |
| | $= 550 \times (0.8/\sin 30 - 0.6\cos 60)$ | M1* | Moment of weight about P |
| | $T = 258$ | A1 | $550 \times (1.6 - 0.3)$ Accept to 2sf |
| | | M1* | Resolving vertically, 3 terms needed |
| | $R = 550 - T\cos 30$ | A1 | Value for $T$ not required |
| | $Fr = T\sin 30$ | B1* | Value for $T$ not required; accept $<$ or $\leq$ |
| | $\mu = 129/326.6$ | M1dep* | For correct use of $F = \mu R, R \neq 550$ |
| | $\mu = 0.395$ | A1 | |
| | | [10] | |
| **OR** | | | |
| | $T \times 0.8/\tan 30 + 550 \times 0.6\cos 60 = R \times 0.8/\cos 60$ | M1* | Moments about V, 3 terms needed |
| | $R = 550 - T\cos 30$ | A2 | A1 for two terms correct |
| | Solve for $T$ or $R$: $T = 258$ or $R = 326.5625$ | M1* | Resolving vertically, 3 terms needed |
| | $Fr = T\sin 30$ | B1* | accept $<$ or $\leq$ |
| | $\mu = 129/326.6$ | M1 | For correct use of $F = \mu R, R \neq 550$ |
| | $\mu = 0.395$ | A1 | Only one needed. Accept to 2sf. For correct use of $F = \mu R, R \neq 550$ |
| | | [10] | |
| **OR** | | | |
| | $Fr \times 1.6\cos 30 + 550 \times (1.6\sin 30 + 0.6\sin 30) = R \times (1.6 + 1.6\sin 30)$ | M1* | Moments about Q, 3 terms needed |
| | $R = 550 - T\cos 30$ | A2 | A1 for two terms correct |
| | $Fr = T\sin 30$ | B1* | accept $<$ or $\leq$ |
| | Solving for at least one of $R$, $Fr$, or $T$: Either $R = 326.5625$, or $Fr = 129(.0017008)$, or $T = 258$ | M1 | |
| | $\mu = 129/326.6$ | M1dep* | |
| | $\mu = 0.395$ | A1 | |
| | | [10] | |7
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{65c47bd2-eace-4fec-b1e6-a0c904c4ec3f-4_474_912_260_493}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A uniform solid cone of height 0.8 m and semi-vertical angle $60 ^ { \circ }$ lies with its curved surface on a horizontal plane. The point $P$ on the circumference of the base is in contact with the plane. $V$ is the vertex of the cone and $P Q$ is a diameter of its base. The weight of the cone is 550 N . A force of magnitude $F \mathrm {~N}$ and line of action $P Q$ is applied to the base of the cone (see Fig. 1). The cone topples about $V$ without sliding.\\
(i) Calculate the least possible value of $F$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{65c47bd2-eace-4fec-b1e6-a0c904c4ec3f-4_528_1143_1302_500}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The force of magnitude $F \mathrm {~N}$ is removed and an increasing force of magnitude $T \mathrm {~N}$ acting upwards in the vertical plane of symmetry of the cone and perpendicular to $P Q$ is applied to the cone at $Q$ (see Fig. 2). The coefficient of friction between the cone and the horizontal plane is $\mu$.\\
(ii) Given that the cone slides before it topples about $P$, calculate the greatest possible value for $\mu$.
\hfill \mbox{\textit{OCR M2 2011 Q7 [14]}}